Area of the Region Problem | AMC-10A, 2007 | Problem 24

Circle centered at \(A\) and \(B\) each have radius \(2\), as shown. Point \(O\) is the midpoint of \(\overline{AB}\), and \(OA = 2\sqrt {2}\). Segments \(OC\) and \(OD\) are tangent to the circles centered at $A$ and $B$, respectively, and $EF$ is a common tangent. What is the area of the shaded region \(ECODF\)?

We have to find out the area of the region \(ECODF\) i.e of gray shaded region.this is not any standard geometrical figure (such as circle,triangle...etc).so we can not find out the value easily.Now if we join \(AC\),\(AE\),\(BD\),\(BF\).Then \(ABFE\) is a rectangle.then we can find out the required area by [ area of rectangle \(ABEF\)- (area of arc \(AEC\)+area of \(\triangle ACO\)+area of \(\triangle BDO\)+ area of arc \(BFD\))]

Can you find out the required area.....?

Given that Circle centered at \(A\) and \(B\) each have radius \(2\) and Point \(O\) is the midpoint of \(\overline{AB}\), and \(OA = 2\sqrt {2}\)

Area of \(ABEF\)=\(2 \times 2 \times 2\sqrt 2\)=\(8\sqrt 2\)

Now \(\triangle{ACO}\) is a right triangle. We know \(AO=2\sqrt{2}\)and \(AC=2\), so \(\triangle{ACO}\) is isosceles, a \(45\)-\(45\) right triangle.\(\overline{CO}\) with length \(2\). The area of \(\triangle{ACO}=\frac{1}{2} \times base \times height=2\). By symmetry, \(\triangle{ACO}\cong\triangle{BDO}\), and so the area of \(\triangle{BDO}\) is also \(2\).now the \(\angle CAO\) = \(\angle DBO\)=\(45^{\circ}\). therefore \(\frac{360}{45}=8\)

So the area of arc \(AEC \) and arc \(BFD\)=\(\frac{1}{8} \times\) area of the circle=\(\frac{\pi 2^2}{8}\)=\(\frac{\pi}{2}\)

can you finish the problem........

Therefore the required area by [ area of rectangle \(ABEF\)- (area of arc \(AEC\)+area of \(\triangle ACO\)+area of \(\triangle BDO\)+ area of arc \(BFD\))]=\(8\sqrt 2-(\frac{\pi}{2}+2+2+\frac{\pi}{2}\))=\(8\sqrt 2 -4-\pi\)

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